Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 5, MAXLOGN = 19, MAXSIZE = 30, inf = 1e9 + 7;
int n, m, q;
int par[MAXN];
int deep[MAXN];
int dp[MAXLOGN][MAXN];
vector<int> G[MAXN];
vector<int> citi[MAXN];
vector<int> path[MAXLOGN][MAXN];
vector<int> ans, k;
void cheak_ans() {
  if (ans.size() > MAXSIZE) {
    k.clear();
    int T = inf;
    for (int j = 0; j < ans.size(); j++)
      if (ans[j] < T) T = ans[j];
    k.push_back(T);
    T = inf;
    for (int i = 1; i < 12; i++) {
      for (int j = 0; j < ans.size(); j++)
        if (ans[j] < T && ans[j] > k.back()) T = ans[j];
      k.push_back(T);
      T = inf;
    }
    ans.clear();
    ans = k;
  }
}
void cheak_path(int a, int b) {
  if (path[a][b].size() > MAXSIZE) {
    k.clear();
    int T = inf;
    for (int j = 0; j < path[a][b].size(); j++)
      if (path[a][b][j] < T) T = path[a][b][j];
    k.push_back(T);
    T = inf;
    for (int i = 1; i < 12; i++) {
      for (int j = 0; j < path[a][b].size(); j++)
        if (path[a][b][j] < T && path[a][b][j] > k.back()) T = path[a][b][j];
      k.push_back(T);
      T = inf;
    }
    path[a][b].clear();
    path[a][b] = k;
  }
}
void cheak_citi(int a) {
  if (citi[a].size() > MAXSIZE) {
    k.clear();
    int T = inf;
    for (int j = 0; j < citi[a].size(); j++)
      if (citi[a][j] < T) T = citi[a][j];
    k.push_back(T);
    T = inf;
    for (int i = 1; i < 12; i++) {
      for (int j = 0; j < citi[a].size(); j++)
        if (citi[a][j] < T && citi[a][j] > k.back()) T = citi[a][j];
      k.push_back(T);
      T = inf;
    }
    citi[a].clear();
    citi[a] = k;
  }
}
void dfs(int v = 0, int p = 0) {
  par[v] = p;
  deep[v] = deep[p] + 1;
  for (int i = 0; i < G[v].size(); i++)
    if (G[v][i] != p) dfs(G[v][i], v);
}
void LCA(int v, int u) {
  ans.clear();
  if (deep[v] < deep[u]) swap(v, u);
  ans.insert(ans.end(), citi[v].begin(), citi[v].end());
  int dis = deep[v] - deep[u];
  for (int i = 0; i < MAXLOGN; i++)
    if ((dis >> i) & 1) {
      ans.insert(ans.end(), path[i][v].begin(), path[i][v].end());
      cheak_ans();
      v = dp[i][v];
    }
  if (u == v) return;
  ans.insert(ans.end(), citi[u].begin(), citi[u].end());
  cheak_ans();
  for (int i = MAXLOGN - 1; ~i; i--)
    if (dp[i][u] != dp[i][v]) {
      ans.insert(ans.end(), path[i][v].begin(), path[i][v].end());
      v = dp[i][v];
      ans.insert(ans.end(), path[i][u].begin(), path[i][u].end());
      u = dp[i][u];
      cheak_ans();
    }
  ans.insert(ans.end(), citi[par[v]].begin(), citi[par[v]].end());
  cheak_ans();
  return;
}
int main() {
  scanf("%d%d%d", &n, &m, &q);
  for (int i = 1; i < n; i++) {
    int v, u;
    scanf("%d%d", &v, &u);
    v--;
    u--;
    G[v].push_back(u);
    G[u].push_back(v);
  }
  dfs();
  for (int i = 0; i < n; i++) G[i].clear();
  for (int i = 1; i <= m; i++) {
    int v;
    scanf("%d", &v);
    citi[--v].push_back(i);
    cheak_citi(v);
  }
  for (int i = 0; i < n; i++) {
    dp[0][i] = par[i];
    path[0][i].insert(path[0][i].end(), citi[par[i]].begin(),
                      citi[par[i]].end());
    cheak_path(0, i);
  }
  for (int i = 1; i < MAXLOGN; i++)
    for (int j = 0; j < n; j++) {
      dp[i][j] = dp[i - 1][dp[i - 1][j]];
      path[i][j] = path[i - 1][j];
      path[i][j].insert(path[i][j].end(), path[i - 1][dp[i - 1][j]].begin(),
                        path[i - 1][dp[i - 1][j]].end());
      cheak_path(i, j);
    }
  while (q--) {
    int v, u, a;
    scanf("%d%d%d", &v, &u, &a);
    v--;
    u--;
    LCA(v, u);
    a = min(a, (int)ans.size());
    sort(ans.begin(), ans.end());
    printf("%d ", a);
    for (int i = 0; i < a; i++) printf("%d ", ans[i]);
    printf("\n");
  }
  return 0;
}